1038 Recover the Smallest Number (30 分)-优快云博客

本文链接：https://blog.youkuaiyun.com/weixin_42582136/article/details/102327178

本文介绍了一个算法问题，即从给定的数字段集合中恢复出可能的最小数字。通过输入一系列数字段，算法将这些段进行排序并组合，形成最小的可能数字。文章提供了完整的C++代码实现，展示了如何通过自定义比较函数对数字段进行排序，以及如何处理前导零的问题。

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Given a collection of number segments, you are supposed to recover the smallest number from them. For example, given { 32, 321, 3214, 0229, 87 }, we can recover many numbers such like 32-321-3214-0229-87 or 0229-32-87-321-3214 with respect to different orders of combinations of these segments, and the smallest number is 0229-321-3214-32-87.

Input Specification:
Each input file contains one test case. Each case gives a positive integer N (≤10
4
) followed by N number segments. Each segment contains a non-negative integer of no more than 8 digits. All the numbers in a line are separated by a space.

Output Specification:
For each test case, print the smallest number in one line. Notice that the first digit must not be zero.

Sample Input:
5 32 321 3214 0229 87
Sample Output:
22932132143287

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int n;
	cin >> n;
	vector<string>re(n);
	for (int i = 0; i < n; i++)
		cin >> re[i];
	sort(re.begin(), re.end(), [](string& a, string& b) {
		return a + b < b + a;});
	string ans;
	for (int i = 0; i < n; i++)
		ans += re[i];
	string ret;
	int i;
	for (i = 0; i < ans.size()&&ans[i] == '0'; i++);
	for (int j=i; j < ans.size(); j++)
		ret += ans[j];
    cout<<(i==ans.size()?"0":ret);
	return 0;
}